Written on 14 September 2020.

Even More Calculus

Sections:

• 0, Chapter 1, Introduction,

  Introduction

  The following units of work on calculus bring together extended methods in both Differentiation and Integration. After studying this unit you should

  • be able to differentiate and integrate basic trigonometric functions
  • understand how to calculate rates of change
  • be able to integrate using substitutions
  • be able to integrate by parts
  • be able to formulate and solve simple first order differential equations.

  The topic is divided into the following 5 sections:

  1. Derivatives of Trigonometric Functions
  2. Rates of Change
  3. Integration by Substitution
  4. Integration by Parts
  5. First Order Differential Equations
• 0, Chapter 2, Derivative of Trigonometric Functions,

  Derivative of Trigonometric Functions

  We start by reminding ourselves of the definition of a derivative of a general function

  The derivative of a general function

  The derivative of a general function  \(f\left(x\right)\) is given by \(\frac{df}{dx} = ^{\lim }_{h \rightarrow 0} \left\{\frac{f\left(x+h\right) -f\left(h\right)  }{h} \right\}\)

  You are going to obtain the derivatives of \(\sin x\), \(\cos x\), and \(\tan x\).

  Remember that the derivative of a function at a given point is given by the gradient of that function at the given point.
  Suppose you look at a sketch of the graph \(\sin x\) and sketch in the tangents for values of \(x\)at the points below
  
    \( -2 \pi\), \(-\frac {3 \pi}{2}\), \( -\pi\), \(-\frac{\pi}{2}\), \(0\), \(\frac{\pi}{2}\), \(\pi\), \(\frac{3 \pi}{2}\), \(2 \pi\)

  This is shown below.

  

  These tangent values when plotted against \(x\) will look like the sketch below.

  

  This suggests that the graph of the function for the tangent may be something like the sketch opposite which suggests that the derivative of \(\sin x\) might be \(\cos x\), but this illustration is certainly not a proof.

  

• 1, Chapter 2.1, Geometric proof,

  Geometric proof

  It is also possible to use a circle of unit radius to show that, for example 
  
    \(\frac{df}{dx} = \left(\sin x\right) = \cos x\) 

  In the diagram shown below \(h\) is small, and 
  
  \(A′ B′ = \sin \left(x + h\right)\) and   \(AB = sin x\)
  
  Hence \(\sin \left(x + h\right) – sin x = A′ P\), and
  
    \(A′ A = radius \times angle = h\)

  

  Now assume that when \(h\) is very small the arc \(A′ A\) is approximately straight.
  
  For the triangle \(A  A′ P\),we can deduce

  \(\frac {\sin \left(x + h\right) - \sin x}{h} = \frac {A′ P}{A′ A} \approx \cos x\)

  So, approximately, if \(h\) is very small
  
  \( \frac{\sin \left(x+h\right) - \sin x}{h} \approx \cos x\)

  And taking the limit as \( h \rightarrow 0\),   \(\frac {d}{dx} \left(\sin x\right) = \cos x\)

  

• 1, Chapter 2.2, Key Results,

  Key Results

  Below are the three crucial derivatives and their inverse integration:

  PUT NEXT RESULTS IN REMEMBER BOX

  \(\frac{d}{dx}\left(\sin x\right)  = \cos x\)   \(\frac{d}{dx}\left(\cos x\right)  = - \sin x\)   \(\frac{d}{dx}\left(\tan x\right)  = - \sec^{2} x\)

  \(\int{\cos x\;dx = \sin x+c } \)   \(\int{\sin x\;dx = -\cos x+c } \)   \(\int{\sec^{2} x\;dx = \tan x+c }\)

  Revision of secant, cotangent and cosecant.

  

  \(\sec \theta = \frac{Hypotenuse}{Adjacent} = \frac{1}{\cos \theta}\)
  \(\cot \theta = \frac{Adjacent}{Opposite} = \frac{1}{\tan \theta} = \frac {\cos \theta}{ \sin \theta}\)
  \(\csc \theta = \frac {Hypotenuse}{Opposite} = \frac{1}{\sin \theta}\)

• 1, Chapter 2.2, Task 1, Worked Example 1,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example

    Find \(\int{x(3x-1)^{2} } \) using a suitable substitution.

    Reveal Solution
  • Worked Example

    Find \(\int_{0}^{1}{x(3x^{2}+2 )^{3}dx } \)

    Reveal Solution
• 1, Chapter 2.3, Trig Revision and Formulae,

  Trigonometric Revision and Formulae

  Key Formulae

  Note that \(\sin^{2}x\) means find \(\sin x\), then square the result.
  
  And \(\sin \left(x^{2}\right)\) means to square \(x\), then the sine function.

  Trig Formulae

  As \(\sin^{2}x + \cos^{2}x =1\) we can deduce
  
  \(\sin^{2} x = 1 - \cos ^{2} x\)
  \(\cos ^{2}x = 1 - \sin ^{2} x\)
  \(\tan^{2}x = \sec^{2}x-1\)
  \(\cot^{2}x= \csc^{2} x -1\)

  Note - throughtout the unit we write cosec\(x\) as csc\(x\)

  Double Angle Formulae
  
  \(\sin 2x= 2 \sin x \cos x\)
  \(\cos 2x = 1 - 2 \sin^{2} x\)
  \(\tan 2x = \frac{2 \tan x}{1 - \tan^{2} x}\)

  Differentiating
  
  

  \(\frac{d}{dx} \left(\frac{1}{\tan x} \right) = \frac{1}{\left(1 + x^{2}\right) } \)
  \(\frac{d}{dx} \left(\frac{1}{\sin x} \right) = \frac{1}{\left(\sqrt{\left(1 -x^{2}\right) } \right) } \)

  \(\frac{d}{dx} \left(\csc x\right) = - \csc x \cot x\)
  \(\frac{d}{dx} \left(\sec x\right) = \sec x \tan x\)
  \(\frac{d}{dx} \left(\cot x\right) = - \csc^{2} x\)

• 0, Chapter 3, Rates of Change,

  Rates of Change

  From your earlier work on differentiation, you will recognise that the rate of change of a variable, say \(f\), is given by \(\frac{df}{dt}\), where \(t\) denotes time. Often though, \(f\) is a function of another variable, say \(x\), so that you need to use the ‘function of a function’ rule,

  PUT NEXT RESULTS IN REMEMBER BOX

  Function of Function Rule

  \(\frac{df}{dt} = \frac{df}{dx} \frac{dx}{dt}\)

• 1, Chapter 3, Task 1, Worked Example 2,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example

    The radius of a circular oil slick is increasing at the rate of \(2\;m/s\)which is \(\frac{dr}{dt}\). Find the rate at which the area of the slick is increasing when its radius is a) \(10\;m\), b) \(50\;m\).

    Reveal Solution
  • Worked Example

    A spherical balloon is blown up so that its volume increases at a constant rate of \(10\;cm^{3}/s (\frac{dV}{dt})\). Find the rate of increase of the radius when the volume of the balloon is \(1000\;cm^{3}\).

    Reveal Solution
• 0, Chapter 4, Integration by Substitution,

  Integration by Substitution

  This method is the equivalent to using a change of variable when differentiating composite functions. In general' 

  If   \(y = \int{f \left(x\right)} dx\)
  
  then   \(\frac{dy}{dx} = f \left(x\right)\)

  We will use this fact to derive a rule for integrating called integration by substitution.

  PUT REMEMBER BOX IN HERE

  If we let \(x = g\left(u\right)\) for some function \(g\) of a new variable \(u\), then \(y\) becomes a function of \(u\) and
  
  \(\frac{dy}{du} = \frac{dy}{dx} \times \frac{dx}{du}\)

  \(\frac{dy}{du} = f \left(x\right) \frac{dx}{du} = f \left(g\left(u\right) \right) \frac{dx}{du} \)

  Then
  
  \(y = \int{f \left(g\left(u\right) \right) \frac{dx}{du}} du\).

• 1, Chapter 4, Task 1, Worked Example 3,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example

    Find \(\int{x(3x-1)^{2} } \) using a suitable substitution.

    Reveal Solution
  • Worked Example

    Find \(\int_{0}^{1}{x(3x^{2}+2 )^{3}dx } \)

    Reveal Solution
• 0, Chapter 5, Integration by Parts,

  Integration by Parts 1

  Our second method of integration is a very important one which is particularly useful for a number of different integrals. 
  You can see how it works by first differentiating with respect to \(x\), \(\frac{d}{dx} \left(x \sin x\right) = x + \cos x\).

  Let's explore an example problem first which illustrates the method

  By assuming \(\int{xe^{x}dx } =(ax+b)e^{x} +c\) find the values of the constants \(a\) and \(b\) so that the result is valid.

  Solution

  We differentiate both sides of the equation above to give:

  \(x e^{x}= \frac{d}{dx}((ax+b)e^{x}+c)\)
  \(= (ax+b)e^{x}ae^{x}+0\)
  \(=(ax+a+b)e^{x}\)

  So comparing coefficients, \(a=1\)and \(a+b=0\) giving \(b=-1\).

  Hence \(\int{xe^{x}dx } =(x-1)e^{x} +c\)

• 1, Chapter 5.1, Integration by Parts 2,

  Integration by Parts 2

  By now you should be wondering if there is a more precise method of evaluating integrals of the type above, where the method requires inspired guesswork to start with. 
  In fact, there is a more formal method

  INSERT A REMEMBER BOX HERE

  If \(u\) and \(v\) are two functions of \(x\) then from your work on differentiation, you know that:

  \(\frac{d}{dx} \left(uv\right) = v \frac{du}{dx} + u\frac{dv}{dx}\)

  If we take the formula above and integrate each side with respect to \(x\), we get 
  
  \(uv =\int{v\frac{du}{dx} } dx + \int{u\frac{dv}{dx} } dx\)

• 1, Chapter 5.1, Task 1, Worked Example 4,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example

    Find \(\int{x \cos x dx}\)

    Reveal Solution
  • Worked Example

    Find \(\int{x e^{x} dx}\)

    Reveal Solution
  • Worked Example

    Find \(\int{lnx dx}\)

    Reveal Solution
  • Worked Example

    Find \(\int_{0}^{\frac{\pi }{2} }{xsinxdx} \)

    Reveal Solution
• 0, Chapter 6, First Order Differential Equations,

  First Order Differential Equations

  You might well wonder why you have studied integration in such depth.  The main reason is the crucial role it plays in solving what are known as differential equations. 
  
  Here you will only be dealing with first order differential equations, which can be expressed, in general, in the form

  \(\frac {dy}{dx} = f \left(x,y\right)\)

  when \(y = f \left(x,y\right)\) is an unknown function of \(x\), and \(f\) is a given function of \(x\) and \(y\).

  Examples of such equations include

  \(\frac {dy}{dx} = xy\)

  \(\frac {dy}{dx} = x + y\)

  \(\frac {dy}{dx} = \frac {x}{y}\)
  

  \(\left( y \neq 0\right)\)

  We will explore solving such equations by demonstration in the worked examples that follow

• 1, Chapter 6, Task 1, Worked Example 5,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example

    For \(x\geq 0\), solve \(\frac{dy}{dx}=\frac{x}{y}\) given that \(y=1\) when \(x=0\)

    Reveal Solution
  • Worked Example

    Solve \(\frac{dy}{dx}=\frac{y}{x}\) given that \(y=1\) when \(x=1\)

    Reveal Solution
• 0, Chapter 7, Summary,

  Summary

  Differentiation of trig functions:

  \(\frac{d}{dx} \left(\sin x\right) =\cos x\)
  \(\frac{d}{dx} \left(\cos x\right) = - \sin x\)
  \(\frac{d}{dx} \left(\tan x\right) =  \sec^{2} x\)
  \(\frac{d}{dx} \left(\cot x\right) = - \csc^{2} x\)
  
  

  Integration of trig functions

  \(\int{\cos x\;dx = \sin x+ c }\)
  \(\int{\sin dx = - \cos x+ c } \)
  \(\int{\sec^{2} dx = \tan x+ c } \)
  \(\int{\csc^{2} x\;dx = - \cos x+ c } \)
  
  

  Function of a function: if \(y = f \left(x\right)\) and \(x = g \left(t\right)\), then
  \(\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}\)

  Integration by substitution: if \(y=\int{f\left(x\right) } dx\) and \(x = g \left(u\right) \), then 
  \(y = \int{f\left(g\left(u\right) \right) } \frac{du}{dx} dx\)

  Integration by parts: if \(n\) and \(v\) are functions of \(x\), then
  \(\int{u\frac{dv}{dx} } dx = uv - \int{v\frac{du}{dx} } dx\)

  A first order differential equation is of the form
  \(\frac{dy}{dx} = f \left(x,y\right)\)
  where \(f\) is a function of \(x\) and \(y\).

Interactive Exercises:

• Even More Calculus , https://www.cimt.org.uk/sif/calculus/c5/index.htm
• Text Exercises, https://www.cimt.org.uk/sif/calculus/c5/text.pdf
• Answers, https://www.cimt.org.uk/sif/calculus/c5/text_ans.pdf

YouTube URL: https://www.youtube.com/watch?v=G0LL8jeO0t4&list=PL8ce1n6mRlCtS0rO_C7CNwtjhyqDVkr8g&index=48

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Written on 14 September 2020.

Further Calculus

Sections:

• 0, Chapter 1, Introduction,

  Introduction

  This unit builds on the earlier work in Calculus. After studying this unit you should:

  • be able to determine stationary points of a function and identify whether they correspond to maximum, minimum values of the function or points of inflection
  • differentiate composite functions
  • integrate standard integrals
  • differentiate products and quotients. 

  This unit is divided into the following five sections:

  1. Rate of Change of the Gradient
  2. Stationary Points
  3. Differentiating Composite Functions
  4. Integration Again
  5. Differentiating Products and Quotients. 
• 0, Chapter 2, Rate of Change of the Gradient,

  Rate of Change of the Gradient

  The sketch opposite shows the graph of

  \(y=x^{3} +3x^{2} -9x-4\)

  

  In the Rates of Change unit we saw that the derivative of this function is given by

  \(\frac{dy}{dx}=3x^{2} +6x-9\)

  

  The gradient function, \(\frac{dy}{dx}\), is also a function of \(x\) and can be differentiated again to give the second differential \(\frac{d^{2} y}{dx^{2} }=6x+6\)

  

• 1, Chapter 2, Task 1, Exercise 1,

  Exercise

  to be added

• 0, Chapter 3, Stationary Points,

  Stationary Points

  In this section we explain how to find stationary points. We dedice the method by looking first at an example.

  \(y=2x^{3} +3x^{2} -12x\)

  This is cubic, and a rough sketch of its graph is shown opposite. It has two stationary points (sometimes called turning points) at which the gradient is zero.

  

  We now need to find the co-ordinates of the stationary points. In the Rates of Change Unit, you saw that the nature of the stationary points can be determined by looking at the gradient on each side of the stationary point. The gradients are shown on the diagram.

  

  You can see above the change in the value of the gradient as it rapidly decreases until it reaches zero at the point of maximum value.

  For a maximum value of a function, note that the gradient is decreasing in value as it passes through the value zero at the stationary point, whereas for a minimum value, this gradient is increasing. Hence:

  Remember

  For stationary points for function \(y(x)\)

  \(\frac{dy}{dx} =0\)

  If \(\frac{d^{2} y}{dx^{2} } <0\) at a stationary point, it corresponds to a maximum value of \(y\)

  If \(\frac{d^{2} y}{dx^{2} } >0\) at a stationary point, it corresponds to a minimum value of \(y\)

  

  Note: that there is no need to draw a graph of a function to find its max/min values.

• 1, Chapter 3, Task 1, Worked Example 1,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example

    Find the maximum and minimum points of the following curve:

    \(y=\frac{1}{4}x^{4}+\frac{1}{3}x ^{3} -6x^{2} +3\)     Hence sketch the curve.

    Reveal Solution
  • Worked Example

    Investigate the stationary points for the graph \(y=x^{3}\)

    Reveal Solution
  • Worked Example

    Show that \(y=x^{4}\) has a minimum at \(x=0\) by checking on the values of the derivative on each side of the stationary point.

    Reveal Solution
• 1, Chapter 3, Task 2, Exercise 2,

  Exercise

  to be added

• 0, Chapter 4, Differentiating Composite Functions,

  Differentiating Composite Functions

  The crucial result needed here is the ‘function of a function’ rule. It can be justified by noting that if \(y\) is a function of \(u\), and \(u\) is a function of \(x\), that is:

  \(y=f(u)\) and \(u=g(x)\)

  then a small change, \(𝛿𝑥\) say, in \(x\) will result in a small change, say \(𝛿𝑢\) say in \(u\), which in turn results in a small change, \(𝛿𝑦\) in \(y\). Now

  \(\frac{\delta y}{\delta x} =\frac{\delta y}{\delta u} \times \frac{\delta u}{\delta x} \)

  and if you let \(\delta x\rightarrow 0\), then \(\delta u\rightarrow 0\) and \(\delta y\rightarrow 0\) and the equation becomes that shown below

  Remember

  If \(y=f(u)\) and \(u=g(x)\) then 

  \(\frac{dy}{dx} =\frac{dy}{du} \times \frac{du}{dx} \)

• 1, Chapter 4, Task 1, Worked Example 2,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example

    Differentiate: \((3x+2)^{2} \)

    Another method, which will be even more useful as the functions get more complicated, is to introduce a new variable \(u\) defined by

    Reveal Solution
  • Worked Example

    Differentiate: \(y=e^{x^{2} } \)

    Reveal Solution
• 1, Chapter 4, Task 2, Exercise 3,

  Exercise

  to be added

• 0, Chapter 5, Integration Again,

  Integration Again

  The results that have been developed in the last section are, as you will see, very useful in integration. For example,

  \(y=\left(x+5\right)^{4}\)

  In general \(y=\left(ax+b\right)^{n}\)

  \(\frac{dy}{dx} = 4 \times 1 \times \left(x+5\right)^{3}\)
  \(\int 4\left(4+5\right)^{3}dx=\left(x+5\right)^{4}+c\)
  \(4 \int \left(x+5\right)^{3} = \left(x+5\right)^{4} + c\)
  \( \int \left(xz+5\right)^{3} dx = \int \frac {\left(x+5\right)^{3}}{an} + k\)

  \(\frac {dy}{dx} = na\left(ax+b\right)^{n-1}\)
  \(\int an\left(ax+b\right)^{n-1}dx = \left(ax+b\right)^{n}+c\)
  \(an \int \left(ax+b\right) ^{n-1} dx = \left(ax+b\right)^{n} + c\)

  \(\int \left(ax+b\right)^{n-1}dx=\frac{\left(ax+b\right)^{n}}{an} +k\)

  or 

  \(\int \left(ax + b\right)^{n}dx= \frac {\left(ax+b\right)^{n+1}}{a \left(n+1\right)}+k\)

• 1, Chapter 5, Task 1, Worked Example 3,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example

    Use the results above to evaluate:

    a)     \(\int{\frac{dx}{(x+1)} } \)

    b)     \(\int{\frac{dx}{1-2x} } \)

    Reveal Solution
  • Worked Example

    Find \(\int{\frac{(3x^{2}+x) }{(2x^{3}+x^{2} )}dx }\)

    Reveal Solution
• 1, Chapter 5.1, More on Integration,

  More on Integration

  Another important method of integration is based on the result that if \(y=lnf(x)\) then defining \(u = f(x)\), \(y = ln u\) and \(\frac{dy}{dx} =\frac{dy}{du} \times \frac{du}{dx} \)

  \(\frac{dy}{dx} =\frac{1}{u} \times f(x)\)

  \(\frac{dy}{dx} =\frac{f'(x)}{f(x)} \)

  \(\int{\frac{f'(x)}{f(x)} } dx=lnf(x)+C\)

• 1, Chapter 5.1, Task 1, Exercise 4,

  Exercise

  to be added

• 0, Chapter 6, Differentiating Products and Quotients,

  Differentiating Products and Quotients

  There are very useful formulae for differentiating both products and quotients. For example, if \(y\) is defined as the product of the functions, \(u\) and \(v\) of \(x\), then

  \(y \left(x\right) = u \left(x\right) v \left(x\right)\)  or \( y = uv\)

  Using the basic definition of a derivative, let \(\delta x\) be a small change in \(𝑥\), then consider
  \(y \left(x+\delta x\right) - y \left(x\right) = u \left(x + \delta x\right) v\left(x+ \delta x\right)  - u \left(x\right) v \left(x\right) = \left[u\left(x+\delta x \right) - u \left(x\right) \right] v \left(x + \delta x \right) + u \left(x\right) \left[v \left(x + \delta x\right) - v \left(x\right) \right]\)

  (the middle two terms cancel out)

  Dividing both sides by \(\delta x\) gives

  \(\frac{y\left(x+\delta x\right) - y\left(x\right)  }{\delta x} = \frac{\left(u\left(x+\delta x\right) - uu\left(x\right)  \right) }{\delta x} = \left(v\left(x+\delta x\right) + u\left(x\right)  \right) \frac{\left(v\left(x+ \delta x\right) - v\left(x\right)  \right) }{\delta x}\)

  and taking the limit as \(\delta x \rightarrow 0\), gives

  \( \frac {dy}{dx} = v \frac {du}{dx} + u \frac {dv}{dx}\)

  Note: There is no need to fully understand the derivation of this formula but it is an important result that you need to be able to use.

• 1, Chapter 6, Task 1, Worked Example 4,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example

    If \(y=x^{2}(x-1)^{2}\) use the Product Rule to find \(\frac{dy}{dx} \).

    Reveal Solution
  • Worked Example

    If \(y=e^{-x} (x+1)\) find \(\frac{dy}{dx} \).

    Reveal Solution
• 1, Chapter 6, Task 2, Worked Example 5,

  Worked Example

  The equivalent formula for a quotient \(y=\frac{u}{v} \) is given below:

  \(\frac{dy}{dx} =\frac{(v\frac{du}{dx}-u\frac{dv}{dx} )}{v^{2} } \)

  Use the slider to explore worked examples.

  • Worked Example

    Differentiate \(y=\frac{x-1}{2x-3}\) with respect to \(x\).

    Reveal Solution
  • Worked Example

    Find the stationary points and sketch the curve of \(\frac{x^{3}}{(1+x)}\).

    Reveal Solution
• 0, Chapter 7, Summary,

  Summary

  Second derivatives

  If \(y\) is a function of \(x\), then \(\frac{d^{2}y}{dx^{2}} = \frac{d}{dx} \left( \frac{dy}{dx}\right)\)
  e.g. If \(y=x^{2}\), \(\frac{dy}{dx}=3x^{2}\), \(\frac{d^{2}y}{dx^{2}} = 6x\)

  Stationary points

  If \(y\) is a function of \(x\), then stationary points are defined by values of \(x\) for which \(\frac {dy}{dx} = 0\) 
  e. g. If \(y= x^{2} + x -6, \frac{dy}{dx} = 2x+1\)
  \(=0\) when \(x = -\frac{1}{2}\)
  so \(y\) has stationary point at \(x= -\frac{1}{2}\)

  Maximum and minimum points

  If \(y\) is a function of \(x\) and \(\frac {dy}{dx} = 0\) at \(x=a\). then it is a
  (i) maximum value of \(y\) if \(\frac {d^{2}y}{dx^{3}} \lt 0\) at \(x = 0\)
  (ii) minimum value of \(y\) if \(frac {d^{2}y}{dx^{2}} \gt 0 \) at \(x=a\)
  e.g. If \(y=x^{2} + x - 6, \frac{dy}{dx} = 2x + 1 = 0\) at \(x = -\frac {1}{2}\)
  \(\frac{d^{2}y}{dx^{2}} = 2 \gt 0\) for any value of \(x\).
  Hence minimum at \(x= -\frac {1}{2}\) (see sketch of function)

  Point of inflection

  If \(y\) is a function of \(x\) such that \(\frac{dy}{dx}= \frac {d^{2}y}{dx^{2}} = 0\) at \(x=a\) but \(\frac{d^{3}y}{dx^{3}} \neq 0\) at \(x=a\), then \(x=a\) is a point of inflection.
  e.g. If \(y=x^{3} - 3x^{2} + 3x -1\)
  \(\frac{dy}{dx} = 3x^{2} - 6x + 3 = 3 \left(x^{2} - 2x + 1\right) = 3 \left(x- 1\right) ^{2} = 0\) at \(x = 1\)
  \(\frac{d^{2} y}{dx^{2}} = 6x - 6 = 0\) at \(x=1\)
  \(\frac {d^{3}y}{dx^{3}} = 6x \neq 0\) at \(x = 1\)
  Hence \(x=1\) is a point of inflection (see sketch below).

  Composite functions

  If \(y = f\left(u\right)\) and \(u=g\left(x\right)\), then
  \(\frac{dy}{dx} = \frac{dy}{du}, \frac{du}{dx}\)

  Note also that \(\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}\)
  e.g. If \(y = 4x + 1, \frac{dy}{dx}=4\)
  Also \(x=\frac{1}{4} \left(y-1\right)\) so \(\frac{dy}{dx} = \frac{1}{4}\), confirming the results above

  Exponential functions

  \(\frac{d}{dx} \left(e^{x}\right) = e^{x}\) and 
  \(\frac{d}{dx} \left(e^{kt}\right)\) for constant \(k\)

  Integration: key results

  \(\int\left(ax+b\right)^{n} dx = \frac{\left(ax+b\right)^{n+1} }{a\left(n+1\right) }  + k\) \(\left(n \neq -1\right)\)
  \(\int\frac{1}{\left(ax+b\right) } dx = \frac{1}{a} \ln \left(ax+b\right) +k\)
  \(\int{\frac{\int′{\left(x\right) } }{\int{\left(x\right) } } } dx = \ln f\left(x\right) +k\)

  Differentiating products and quotients

  If \(y \left(x\right) = u\left(x\right) v\left(x\right)\), then
  \(\frac{dy}{dx} = \frac{du}{dx} v + u \frac{dv}{dx}\)
  e.g. \(y = x^{2} \sin x\), then \(u=x^{2}\), \(v=\sin x\) and \(\frac{dy}{dx} = 2x \sin x + x^{2} \cos x\)

  If \(y\left(x\right) = \frac {u \left(v\right)}{v \left(x\right)}\), then \(\frac{dy}{dx} =\frac{\frac{du}{dx}v - u\frac{dv}{dx}  }{v^{2}} \)
  e.g. \(y = \frac {x^{2}}{1 -x}, u=x^{2}, v=1-x\)
  \(\frac {dy}{dx} = \frac{2x\left(1-x\right) -x^{2}\left(-1\right) }{\left(1-x\right) ^{2}} = \frac{-x^{2} +2x}{\left(1-x\right)^{2}} \)

Interactive Exercises:

• Further Calculus Questions, https://www.cimt.org.uk/sif/calculus/c4/text.pdf
• Further Calculus Answers, https://www.cimt.org.uk/sif/calculus/c4/text_ans.pdf

YouTube URL: https://youtu.be/MvWeCParTrc

File Attachments: media/course-resources/Further-Calculus_Essential-Informations.pdfmedia/course-resources/Further-Calculus_Learning-Objectives.pdfmedia/course-resources/Further-Calculus_PowerPoint-Presentation.pptxmedia/course-resources/Further-Calculus_Text.pdfmedia/course-resources/Further-Calculus_Answers.pdf

Written on 15 September 2020.

Growth and Decay

Sections:

• 0, Chapter 1, Introduction,

  Introduction

  In this unit we focus on practical problems that illustrate growth and decay, introducing the
  exponential function, which has widespread applications. After completing this unit you should

  • appreciate the characteristics of exponential functions of the form \(y=a^{x}\)
  • be able to use the methodology of Carbon Dating to date artifacts
  • understand the concept and characteristics of the exponential function \(y= e^{x}\)
  • appreciate that ln \(x\) is the inverse of \(e^{x}\)
  • be able to solve exponential equations
  • understand the properties of logarithm

  The unit is divided into the following sixsections:

  1. Growth and Decay
  2. Modelling Population
  3. Models of Growth and Decay
  4. Carbon Dating
  5. Rate of Growth
  6. Solving Exponential Equations
  7. Problems of Logarithms

   

• 0, Chapter 2, Modelling Populations,

  Modelling Populations

  Amoebae reproduce by fission, or splitting in two after a certain time. The "parent" cell divides into two smaller copies of itself. The nucleus also divides into two.

  Radioactive substances have 'half lives' which are determined by the time it takes the radioactivity to halve. 

  In 2020 the world was in the grips of a COVID-19 pandemic caused by severe acute respiratory syndrome coronavirus 2.

  These are examples of systems which are modelled by what we call 'exponential' functions.

  The graph below shows the growth in total COVID-19 cases day-to-day for several countries.

  

• 1, Chapter 2, Task 1, Worked Example 1,

  Worked Example

  In the example below we form a model to describe the population growth and then use it to make predictions about future populations.

  • Worked Example

    The world population at the end of 1989 was approximately \(4.5\) billion. If the population grew each year by \(3\%\), what would be the population at the end of 1990, 1991 and 1992?

    Reveal Solution
• 1, Chapter 2.1, Modelling Equations,

  Modelling Equations

  The model used in the previous example can be represented by the equation:

  \(P \left(t\right) = P_{0} a^{t}\)

  where \(P\) is the population at the end of year number \(t, P_{0}\) is the initial population at year \(t = 0\) and \(a\) is a constant.

  So for the world population described above  

  \(P \left(t\right) = 4.5 \; billion \times 1.03^{t}\) for \(t = 1, 2, 3\)

  Equations of the form above can also be used to describe populations that are declining so the constant \(a\) would be less than \(1\).Some species are endangered because they have declining populations for various reasons such as hunting, habitat destruction, new predators or infertility.Models are made to help predict future trends in fish stock levels, which take into account many features like fishing techniques and  environmental conditions.

• 1, Chapter 2.2, TBA 1,

  TBA

  to be added

• 0, Chapter 3, Models of Growth and Decay,

  Models of Growth and Decay

  The previous models show how mathematics can be used to model growth and decay of populations.  Another example is bacteria, which divide in two every minute. 
  
  The growth in numbers is illustrated in the table below.

  

  The table shows a pattern for the number of bacteria at any time after the bacteria was introduced. 
  
  From this pattern, it is easy to see that after t minutes the number of bacteria will be \(2^{t}\).

• 1, Chapter 3, Task 1, Model Example 1,

  Model Example

  We can deduce from our example of bacteria growth that,  \(n = 2^{t}\) is a model for the bacteria growth. The graph for the function is shown here, for the values of \(t\) between \(0\) and \(4\).

  

  Drawing a smooth curve through these discrete points gives a continuous model of growth.

  Many situations which have models like this have continuous not discrete domains, so the curve is typical of this type of function.

   

   

• 1, Chapter 3.1, Exponential Function,

  Exponential Function

  Any function of the form \(a^{x}\) , where \(a\) is a positive constant is called an exponential function (as \(x\) is the 'exponent' or power of \(a\)) . 
  
  Although the example above was only defined for positive values of the exponent \(\left( t \geq 0 \right)\), exponential functions are defined for any real value.

  For instance, if \(f \left(x\right) = 3^{x}\), then \(f\left(-1\right) = 3^{-1} = \frac {1}{3}\)

• 1, Chapter 3.1, Task 1, Worked Example 2,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example

    a)     Use a graphic calculator or computer to help you make sketches of these functions, i) to iv) below, using the same pair of axes. Use a range of the values between \(−3\) and \(+3\), and \(0\) to \(+30\) on the \(y-axis\).

    i)   \(y=3^{x}\)
    ii)   \(y=2^{x}\)
    iii)   \(y=1^{x}\)
    iv)   \(y=(0.5)^{x}\)

    b)     Do the curves have a common point?

    c)     What is the relationship between \(y=2^{x}\) and \(y=0.5^{x}\) ?

    d)     What happens as \(x\) becomes large and positive?

    Reveal Solution
  • Worked Example

    Draw the graphs of \(y=3^{-x}\) and \(y=2^{-x}\).

    Compare them with \(y=3^{x}\) and \(y=2^{x}\).

    Describe the behaviour for large positive values of \(x\).

    Reveal Solution
  • Worked Example

    The government’s statistical service made several predictions about the United Kingdom’s population in 2014. One of these was that the population will grow by \(2\%\) every year. If the population in 2014 was \(68\;million\), form a model for the UK’s future population. Use it to draw a graph or spreadsheet of the projected population, and from this estimate the year when the population will equal \(80\;million\).

    Reveal Solution
• 1, Chapter 3.2, TBA 2,

  TBA

  to be added

• 0, Chapter 4, Carbon Dating,

  Carbon Dating

  Carbon 14 (\(14_{C}\)) is an isotope of carbon with a half life of 5730 years.  It exists in the carbon dioxide in the atmosphere, and all living things absorb some Carbon 14 as they breathe. 

  This remains in an animal or plant, and is constantly added to until the organism dies. 

  After death, the Carbon 14 decays, reducing to half the amount stored in the body after 5730 years. 

  The amount halves again after another 5730 years, and so on, with no new Carbon 14 absorbed.

  In 1946 an American scientist, Willard Libby, developed a way of 'dating' archaeological objects by measuring the Carbon 14 radiation present in them. 

  This radioactivity is compared with that found in things living now.

  For instance, if bones of recently dead animals produce 10 becquerels per gram of bone carbon (a becquerel ( or bec)) is the unit of radioactivity), and an old bone produces only 5 becquerels, the radioactivity has halved since the animal which had the old bone died.

  As the half life of Carbon 14 is 5730 years, this would mean the animal died in 3715 BC approximately, as tested in 2015.

• 1, Chapter 4.1, TBA 3,

  TBA

  to be added

• 0, Chapter 5, Rate of Growth,

  Rate of Growth

  Suppose a colony of bacteria doubles in number every minute as every member of the colony divides in two.  So if there are 2 bacteria at the start of the colony, there will be \(4\) a minute later (an increase of \(2\) in one minute), \(8\) two minutes later (an increase of \(4\) in one minute) and so on. 

  As the number of bacteria increases, so the rate at which that number increases goes up.  So the rate of increase of an exponential function is closely related to the value of the function at any point. 

  This suggests that exponential functions and their derivatives are closely linked as we will see in the next example.

• 1, Chapter 5, Task 1, Worked Example 3,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example

    Plot and draw the following curves: a)   \(y=2^{x}\)       b)   \(y=3^{x}\)

    Using a ruler to draw tangents to each of the curves, calculate the gradient of each curve at \(x=0, 0.5, 1, 1.5\) and \(2\).

    Plot the values for the gradients for each graph.

    Are the graphs related?

    Reveal Solution
  • Worked Example

    Compare the following graphs with the graph \(y=e^{x}\).

    i)     \(y=e^{x}\)

    

     

    ii)     \(y=e^{(x+1)}\)

    

     

    iii)     \(y=e^{(x-2)}\)

    

     

    iv)     \(y=e^{x}+1\)

    

     

    v)     \(y=e^{-x}\)

    

    Reveal Solution
  • Worked Example

    Find \(x\) if \(x^{x}=100\). Give your answer to two decimal places.

    Reveal Solution
  • Worked Example

    Solve, to 3 significant figures. the equation \(3e^{2x-1}=5\).

    Reveal Solution
• 1, Chapter 5.1, TBA 4,

  TBA

  to be added

• 0, Chapter 6, Solving Exponential Equations,

  Solving Exponential Equations

  Earlier in this unit, we used exponential functions as models, and then graphs to estimate the solution to a problem. 

  The logarithmic function allows you to calculate rather than estimate these solutions. Lets see how this is done by way of example.

• 1, Chapter 6, Task 1, Worked Example 4,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example

    A bacteria colony doubles in number every minute, from a starting population of one. The population model is \(P(m)=2^{m}\) where \(P\) is the population and \(m\) the number of minutes since the colony was started. Find the time when the population first equals \(1000\).

    Reveal Solution
  • Worked Example

    Solve \(3^{2x-1}=5^{x}\), giving your answer to \(2\) decimal places.

    Reveal Solution
• 1, Chapter 6.1, TBA 5,

  TBA

  to be added

• 0, Chapter 7, Properties of Logarithms,

  Properties of Logarithms

  Remember

  As well as obeying the rule

  \(\ln \left(a^{x}\right) = x \ln a\)

  logarithms also obey, for any real numbers \(a, b,\)

  \(\ln \left(ab\right) = \ln a+ \ln b\)
  and
  \( \ln \frac{a}{b} = \ln a - \ln b\)

• 1, Chapter 7, Task 1, Worked Example 5,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example

    The relationship between length, l (cm), and frequency, f (hz), can be found experimentally. The next table shows some data collected by experiment for a particular type of string.

    \(f=al^{b}\)

    

    Use logarithms to 'linearise' the relationship.

    Plot \(ln f\) on a vertical axis and \(ln l\) on the horizontal, and draw a line of best fit.

    Find the gradient and intercept with the vertical axis of this line, and so determine the values of \(a\) and \(b\).

    Reveal Solution
  • Worked Example

    For the data above, plot a graph of \(lnT\) against \(lnR\), and use it to estimate the values of the constants \(a\) and \(b\).

    \(T=aR^{b}\)

    Reveal Solution
• 1, Chapter 7.1, Exponential (power) Function,

  Exponential (power) Function

  The graphs of the exponential (power) function, \(y = ba^{2}\) , for constants \(a\) and \(b\), is given by (for \(b \gt 0\)):

  

  If \(y = e^{x}, \frac{dy}{dx}=e^{x}\)

  If \(e^{x} = a\), then \(x = \ln a\)

  \( y = e^{x}\) and \(y = \ln x\) are inverse functions

  Properties of logarithms:

  \(\ln \left(a^{x}\right) = x \ln a\)

  \(\ln \left(ab\right) = \ln a + \ln b\)

  \(\ln \left(\frac {a}{b} \right) = \ln a - \ln b\)

Interactive Exercises:

• Growth and Decay Questions, https://www.cimt.org.uk/sif/calculus/c3/text.pdf
• Growth and Decay Answers, https://www.cimt.org.uk/sif/calculus/c3/text_ans.pdf

YouTube URL: https://youtu.be/46cQAVDVPd8

File Attachments: media/course-resources/Growth-Decay_Learning-Objectives.pdfmedia/course-resources/Growth-Decay_PowerPoint-Presentation.pptxmedia/course-resources/Growth_Decay_Text.pdfmedia/course-resources/Growth-Decay_Answers.pdfmedia/course-resources/Growth-Decay_Essential-Information.pdf

Written on 15 September 2020.

Integration

Sections:

• 0, Chapter 1, Introduction,

  Introduction

  Integration is the reverse of differentiation so you need to be familiar with basic concepts that underpin methods of differentiation.

  After studying this unit you should

  • understand the concept of integration and relate it to finding the area under a graph
  • be familiar with the general formula for the area under a straight line graph between set limits
  •  appreciate the linking between differentiation and integration and be able to integrate powers and exponentials
  • understand the difference between indefinite integrals and integrals with limits
  • be able to apply integration with initial limits to problems in context.

  The unit  of work on Integration is divided into the following six sections.

  1. Estimating Future Population
  2. Notation for Area
  3. General Formula
  4. The Reverse of Differentiation
  5. Finding Areas
  6. Initial Conditions

   

• 0, Chapter 2, Estimating Future Population,

  Estimating Future Population

  What is integration?

  Integration cna be thought of  the process of finding the area under a graph. An example of an area that integration can be used to calculate is the shaded one shown in the diagram.

  There are several ways of estimating the area - this unit includes a brief look at such methods - but the main objective is to discover a way to find the area exactly.

  

  Finding the area under a graph is not just important for its own sake. There are a number of problems in science and technology that need integration for solutions to practical problems.

  The next section starts by looking at an example of such a problem.

• 1, Chapter 2.1, Estimating Future Populations,

  Estimating Future Populations

  Shortly after the Second World War, it was decided to establish several ‘new towns’. Two well-known examples are

  • Crawley, near London,
  • Newton Aycliffe in the north-east

  In these cases the ‘new towns’ were based on existing small communities.

  The city of Milton Keynes though was built up on a completely new site, and has continued to expand.

  Clearly such ambitious developments require careful planning and one factor that needs to be considered is population growth. Apart from anything else, the services and infrastructure of a new community need to be planned taking into account the projected population increase: chaos would ensue if, for instance, there were 5000 children but only enough schools for 3000!

  Imagine you are planning a new town. You have been advised that planned population growth will conform to the following model:

  Initial growth rate 6000 people per year; thereafter the rate of growth will decrease by 30% every five years.

  The task is to estimate what the total population will be 30 years later. One approach to this problem is shown in the next section.

   

• 1, Chapter 2.1, Task 1, Example 1,

  Example 

  We start by setting up a mathematical model. Let t stand for the time in years; when \(t  = 0\) the growth rate is \(6000\) people per year.

  When  \(t  = 5\) (after \(5\) years) the growth rate is \(6000\) less \(30\%\),
  
  i.e.  \(0.7 \times 6000 = 4200\).

  

  A very rough estimate can be obtained by making the assumption that the growth rate remains fixed at \(6000\) throughout the first \(5\) years.
  
  Then after 5 years the population will be:
  
  \( 5 \times 6000 = 30 \; 000 \)
  
  Now suppose the growth rate over the next five years is a constant \(4200\) per year.
  
  After 10 years the population will be: \(30 \; 000 + 5 \times 4200 = 51 \; 000\)
  
  Continuing this process we obtain an estimate for the population after 30 years by calculating:
  
  \(5 \times (6000 + 4200 + 2940 + 2058 + 1441 + 1008 + 708) = 5 \times 18355 =91775\)

  This figure will clearly be an over-estimate.

  

  A similar process can be applied to give an under-estimate.

  In this case we assume the lower growth for a period is equivalent to the rate we used in the following period when we obtained our over-estimate. This will become clearer when you look at the calculation:
  
  First, suppose the growth rate over the first five years is fixed at \(4200\).
  
  After five years the population will be:\( 5 \times 4200 = 21 \; 000 \)
  
  After 10 years the population will be \(21 \; 000 + 5 \times 2940 = 35 \; 700\)
  
  Continuing this process we obtain a second estimate as:
  
  \( 5 \times (4200 + 2940 + 2058 + 1441 + 1008 + 708 + 494) = 5 \times 12 \; 849 = 64245\)

  We can use these two estimates to make a third, more realistic estimate of the population after 30 years’ growth, taking the average of the two values:
  
  \( (91775 + 64245) \div 2 = 78 \; 010 \)

  

  The two previous diagrams should give a good idea of what the processes we have just completed actually means. We essentially found the area under two histograms; one estimate was too large, the other too small, with the ‘true’ answer lying somewhere in the middle.

  The graph of growth rate against time will not, in truth, be a ‘histogram’ at all. More  realistically, it will resemble the curve shown below on the left, which shows the growth rate decreasing continuously.

  

  The figure on the right shows the same curve with the two histograms superimposed.

  

  The ‘true’ estimate will be the exact area under the curve above.

• 0, Chapter 3, Area under a Graph,

  Area under a Graph

  Three examples have now been encountered in which it has been important to find the area under a graph.

  Mathematicians have for many centuries appreciated the importance of areas under curves.

  The process of working out such areas is called ‘integration’ and early approaches to the problem were similar to the method investigated so far; namely, the splitting of areas into several thin rectangles.

  

• 1, Chapter 3.1, Notation for Area under the graph,

  Notation for Area under the graph

  The notation developed by the German mathematician, Leibnitz, quickly became widely adopted and is still used today.

  For example, the area under the curve \(y=4x^{3} - 15x^{2} + 12x + 5\) between \(x=1\) and \(x=2\) is written as

  \(\int _{1}^{2}\left(4x^{2} - 15x^{2} + 12x + 5\right) dx\)

  It is read as the integral between \(1\) and \(2\) of \(4x^{3} - 15x^{2} + 12x +5\)

  The  \(\int\) sign, known as the integral sign, derives from the ancient form of the letter S, for sum.

  The ‘\(dx\)’ represents the width of the small rectangles as they get smaller and smaller.

  The notation above denotes the sum of many  very thin rectangles as the width tends to zero between the limits \(𝑥 = 1\) and \(𝑥 = 2\).

  

• 1, Chapter 3.2, Area under a straight line,

  Area under a straight line

  The activities in the previous section highlighted the fact that, so far, the methods we have used to calculate areas have been estimates.

  When the graph is a straight line, however, it is a simple matter to work out the area exactly.

  Consider the case of a stone thrown vertically downwards from a cliff-top. If its initial velocity is \(5\;metres\) per second then its velocity, v metres per second after t seconds, will be given approximately by the formula

  \(v = 5+10t\)

  How far will it have travelled after \(5\;seconds\), assuming it hasn’t hit anything by then?

  The answer to this question can be worked out by finding the area under, that is, integrating, the velocity-time graph.

  The problem requires finding the shaded area in the diagram below.

  

  Using the integral sign, this area can be written

  \(\int^{2}_{1}\left(5+10t\right) dt\)

  (Note that dt is used here rather than dx as the variable along the horizontal axis is t.)

  The shaded area is a trapezium, the area of which can be worked out using the general rule

  \(Area=\frac{1}{2}\times\left(sum\;of\;parallel\;sides\right) \times \left(distance\;between\;them\right)\)

  In this case the calculation yields \(\frac{\left(5+55\right) \times 5}{2} = 150\) meters

• 1, Chapter 3.2, Task 1, Worked Example 1,

  Worked Example

  • Worked Example 

    Calculate the integral \( \int_{4}^{6}{(20-3x)dx} \)

    Reveal Solution
• 0, Chapter 4, General Formula,

  General Formula 

  It is straightforward to find the area under any straight line graph, say

  \(y=mx+c\)

  between \(x=1\) and \(x=b\), as shown in the figure below.

  Find the value of \(y\) for \(a\) and \(b\); that is 

  \(y=ma+c\) and \(y=mb+c\)

  

  The area between \(a\) and \(b\) is given by

  \(\begin{align*} Area&= \frac {\left(ma+c+mb+c\right)\left(b - a\right)}{2}\\ &=\frac{\left(ma+mb+2c\right)\left(b-a\right)}{2}\\ &=\frac{mb^{2} + mab + 2cb - mab - ma^{2} - 2ca}{2}\\ &= \frac {1}{2} \left(mb^{2} + cb \right) - \frac {1}{2} \left(ma^{2} + ca\right)\\ &= \left[\frac{1}{2} mx^{2} + cx \right]^{b}_{a} \end{align*}\)

  Note the convention used here for limits with the square brackets.

  The function \(\left(\frac{1}{2} mx^{2} + cx\right)\) can be used to find areas efficiently.

  It is called an indefinite integral of the function  𝑚𝑥 + 𝑐, since in this form it has no limits. One way of thinking of it is as the ‘area function’ for the graph of  𝑦 = 𝑚𝑥 + 𝑐. 

  For the moment the ‘area function’, or indefinite integral, will be denoted by A(x), whereas the area between 𝑥 = 𝑎 and 𝑥 = 𝑏 is given by

  \(Area=\int^{b}_{a}\left(mx + c\right) dx = \left[\frac{1}{2} mx^{2} + cx\right] ^{b}_{a}\)

  For the straight line function y = mx + c, this is called a definite integral.

• 1, Chapter 4, Task 1, Worked Example 2,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example 

    Find the area under \( y=3x-5 \) between \( x=2 \) and \( x=10 \).

    Reveal Solution
  • Worked Example 

    Determine the area under the graph of \( y=12-8x \) between \(x=1\) and \(x=3\).

    Reveal Solution
• 0, Chapter 5, The Reverse of Differentiation,

   The Reverse of Differentiation

  Now we move on to search for a method of integrating more difficult functions. For straight lines, the formula of the line and its indefinite integral are connected by the following rule.

  

  Let us demonstrate this rule by looking at one of our previous examples. In the previous section, the equation of the line was 𝑦 = 3𝑥 − 5. The corresponding indefinite integral was \(A \left(x\right) = \frac {3x^{2} }{2} - 5x\) 

  Hence \(\frac{dA}{dx} = 3x - 5 = y\)

  This shows that when we differentiate the answer obtained from an integration we arrive at the original function.

  Remember

  Integration is the reverse of differentation

  In general the area under a graph showing the rate of change of some quantity will give the quantity itself. 

  

• 1, Chapter 5, Task 1, Worked Example 3,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example 

    Calculate these indefinite integrals

    a)   \( \int{x^{2}dx} \)

    b)   \( \int{x^{4}dx} \)

    c)   \( \int{x^{5}dx} \)

    d)   \( \int{x^{-2}dx} \)

    e)   \( \int{x^{-3}dx} \)

    f)   \( \int{x^{-4}dx} \)

    Reveal Solution
  • Worked Example 

    Determine the following indefinite integrals.

    a)   \( \int{5x^{3}dx} \)

    b)   \( \int{ \frac{7}{x^{3}} dx} \)

    c)   \( \int{ (3x^{5} - \frac{2}{x}) dx} \)

    d)   \( \int{ \frac{x+x^{2}}{3} dx} \)

    Reveal Solution
• 0, Chapter 6, Finding Areas,

  Finding Areas

  The original aim of this unit was to calculate exact values of areas under curves. The groundwork for this has now been done.
  
  The procedure is best explained through a worked example but, before you read through it, you might like to remind yourself how indefinite integrals were used to work out areas under straight lines in Section 2

• 1, Chapter 6, Task 1, Worked Examples 4,

  Worked Examples

  Use the slider to explore worked examples.

  • Worked Example 

    Calculate the area under the curve \( y=x^{2} + 2 \) between \(x=1\) and \(x=4\).

    Reveal Solution
  • Worked Example 

    Work out the area under the graph of \( y=10e^{x} + 3x \) between \(x=-1\) and \(x=3\), to one decimal place.

    Reveal Solution
  • Worked Example 

    Evaluate: \( \int_{0}^{6}{(x^{2} - 2x -8)dx} \)

    Reveal Solution
• 0, Chapter 7, Initial Conditions,

  Initial Conditions

  We will see how important the initial conditions are in applications by looking at the worked example in the next section.

• 0, Chapter 8, Summary,

  Summary

  The area under a straight line graph, \(y=mx+c\) between \(x=a\) and \(x=b\) is given by

  \(Area = \left(\frac{1}{2} mb^{2} + cb \right) - \left(\frac {1}{2} ma^{2} + ca\right)\)

  

  Integration is the reverse of differentation

  Standard integrals are

  \(\int x^{n} dx = \frac {x^{n+1}}{n+1} + K\) where \(K\) is constant and \(n\) any integer except \(n=-1\)

  \(\int x^{-1} dx = \ln x + K\) where \(K\) is a constant

  \(\int e^{x} dx = e^{x} + K\)

  Note that

  \(\int\left(au\left(x\right)  + bv\left(x\right) \right)dx = a\)

  \(\int u\left(x\right) dx + b\)

  \int v\left(x\right) dx\)

  where \(a\) and \(b\) are constants, and \(u\) and \(v\) functions of \(x\)

  Area under the curve \(y = f \left(x\right)\) from \(x=a\) to \(x=b\) is given by the integral 

  \(A= \int ^{ b}_{a} f \left(x\right) dx\)

  

   

   

Interactive Exercises:

• Integration Questions, https://www.cimt.org.uk/sif/calculus/c2/text.pdf
• Integration Answers, https://www.cimt.org.uk/sif/calculus/c2/text_ans.pdf

YouTube URL: https://youtu.be/XsnJH7FCPcU

Written on 15 September 2020.

Rates of Change

Sections:

• 0, Chapter 1, Introduction,

  Introduction

  In this unit we begin the development of an important branch of mathematics, namely calculus, based on consideration of finding rates of change. After completing this unit you should

  • be able to find the gradient of curves by measurement
  • understand the difference between instantaneous speed and average speed
  • be able to use straightforward differentiation to find gradients of curves
  • be able to solve optimisation problems based on differentiation.

  This unit is divided into the following seven sections:

  1. Introduction
  2. Instantaneous Speed
  3. Finding the Gradient
  4. Gradient of Quadratics
  5. Differentiation
  6. Optimisation
  7. Real Problems

• 1, Chapter 1.1, Examples of Change,

  Examples of Change

  There are countless examples of change: the thickness of the ozone layer is changing with time; the diameter of a metal ring changes with temperature; the air pressure on a mountain changes with altitude.  In many cases, however, what is important is not whether things change, but how fast they change.

  A very practical problem is in motion where the rate of change of distance with time is speed (velocity) and the rate of change of speed/velocity with time is acceleration.

  Lets look at some examples of rates of change in motion.

• 1, Chapter 1.1, Task 1, Worked Example 1,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example 

    The town of Dorchester in Dorset is \(2\;km\) from end to end, and a \(30\;mph\) speed limit is in force throughout. Although the A35 road now by-passes the town, many drivers consider it quicker, late at night when traffic is light, to drive through the centre of town.

    a)     A driver took \(2\;minutes\;40\;seconds\) to drive through the town. Assuming the car stayed at a constant speed, was the speed limit broken?

    b)     What is the shortest time a driver can take to drive through Dorchester and not break the speed limit?

    Note:

    \(1km \approx 0.6214 miles\)

    \(1 mile \approx 1.6093 km\)

    Reveal Solution
  • Worked Example 

    The graph opposite is a distance-time graph for a car being driven through Dorchester.

    

    a)   Travelling through Dorchester the driver negotiated a major roundabout. Where do you think it is, and how can you tell?

    b)   What was the car’s average speed, in mph, between:
    i)   Grey’s Bridge and Night Club
    ii)   Night Club and Hospital?

    Reveal Solution
• 0, Chapter 2, Instantaneous Speed,

  Instantaneous Speed

  Another method of finding the speed is to use a ‘radar gun’, which is focused on the car as it passes. 
  This gives the instantaneous speed of the vehicle, similar to that shown on the speedometer of the car.

  On a distance-time graph, the instantaneous speed is indicated by the steepness, or gradient, but when the graph is a complicated curve the gradient is difficult to pin down accurately.  One way is to draw a tangent to the curve and to work out the gradient ( Note you will recall that the tangent to a curve at point is the line which' just touches' the curve at that point)

  There are many other examples of a graph where the gradient, or steepness, is useful. This property of the gradient has important applications to all kinds of graphs, as illustrated below.

  

  

  The graph below shows how the air temperature changes with height above sea level. 

  The gradient of the tangent at \(P\) is \(3^{\circ} C/km\). 

  This indicates that,\(1\;km\) above \(P\), the temperature will be approximately \(3^{\circ}C\) higher. 

  The rate of change of temperature with distance is sometimes called the temperature gradient.

  

• 1, Chapter 2, task 1, Worked Example 2,

  Worked Example

  • Worked Example

    In a game of rugby, imagine that a try has been scored, the ball being grounded on one side of the in-goal area. A kick at goal now follows.

    The angle in which the ball must be kicked is denoted by \(\alpha \) ; the smaller the angle, the trickier the kick. The angle depends on how far back the kick is taken. The way it changes is shown in the table.

    

    a)     Draw a graph of angle against distance.

    b)     Estimate the gradient of the graph at \(x=5\;m\), \(15\;m\) and \(40\;m\). Interpret your answers.

    c)     What is the best point from which to take the kick? What is the gradient at this point?

    Reveal Solution
• , , ,

  To be added later

• 0, Chapter 3, Finding the Gradient,

  Finding the Gradient

  If you know the equation of a line it is possible to calculate the gradient of the line without drawing a diagram. We will show this with some examples that follow but here is the general rule.

  Remember

  For any equation of a straight line given in the form \(y=mx+c\)  where \(m\) and \(c\) are constants,

  the gradient of the line is \(m\)

• 1, Chapter 3, Task 1, Worked Example 3,

  Worked Examples

  • Worked Example

    For the function \(y=3x\) find the gradient of the line.

    Reveal Solution
  • Worked Example

    For the function \(y=2x+1\) find the gradient of the line.

    Reveal Solution
• 1, Chapter 3.1, Finding the gradient 2,

  Finding the gradient 2

  If you know the equation of a curve it is possible to calculate the gradient at a point without having to draw tangents to the curve at the point by looking at values around the point. This is best understood by studying the following worked example.

  • Worked Example

    For the function \(y=x^{2}+x-6\), find the values of \(y\) when \(x=0.95\) and \(x=1.05\). Use these values to estimate the gradient at point \(\left(1, −4\right)\).

    Reveal Solution
• , , ,

  to be added

• 0, Chapter 4, Gradient of Quadratics,

  Gradient of Quadratics

  Using the results from Section 3 it is possible to draw the following conclusions:

  • the gradient of \(y=x^{2}\) is given by \(2x\), so the gradient of \(y=ax^{2}\) is \(2ax\)
  • the gradient of \(y=bx\), a straight line, is just \(b\)
  • adding \(a\) constant, \(c\), merely moves the curve or line up or down and does not alter the gradient.

  The general equation of a quadratic is

  \(y = ax^{2} + bx + c\)

  Hence, applying the rules we just deduced, the gradient of a quadratic will be in the form \(2ax + b\).

  Lets look at this rule in some worked examples.

• 1, Chapter 4, Task 1, Worked Example 4,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example

    Find the formula for the gradient of the curve \(y=5x^{2}+7x+10\)

    Reveal Solution
  • Worked Example

    Find the formula for the gradient of the curve \((2x+4)(3x+2)\)

    Reveal Solution
  • Worked Example

    If \(y= \frac{x^{2}-7x}{10}\), what is the gradient when \(x=15\)

    Reveal Solution
• , , ,

  to be added

• 0, Chapter 5, Differentiation,

  Differentiation

  The process of finding ‘gradient functions’ is called differentiation. Another name for the gradient function is the derivative or derived function.

  \(6x^{2} - 8x +9\) is differentiated to give  \(12x - 8\)

  \(12x - 8\) is known as the derivative of \(6x^{2} - 8x + 9\)

  The inventor of this technique is generally thought to have been the English physicist and mathematician Sir Isaac Newton (1643-1747), who developed it in order to explain the movement of stars and planets. However, the German mathematician Gottfried Wilhelm Leibniz (1646-1716) ran him close, and it was Leibniz who was the first actually to publish the idea, in the year 1684. Much vigorous and acrimonious discussion ensued as to who discovered the technique first. Today both are saluted for their genius.

  The notation used by Leibniz is still used today.

  Remember

  The gradient of a straight line is

  \(\frac{change\;in\;y}{change\;in\;x}\)

  Which Leibniz shortened to \(\frac{dy}{dx}\) (read as 'dy by dx').

  For example,   if  \(y = 6x^{2}-8x + 12\)

  \(\frac{dy}{dx} = 12x -8\)

  or this is sometines written as

  \(\frac{d}{dx} \left(6x^{2} - 8x + 12\right) = 12x -8\) 

  \(\frac{d}{dx}\) is the derivative with respect to \(x\)

  or the following notation is also used 

  \(f \left(x\right) = 6 x^{2} - 8x + 12\)

  \(f'\left(x\right) = 12x -8\) (derived function)                

• 1, Chapter 5, Task 1, Worked Example 5,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example 

    a)   \( y=x^{2} \)

    b)   \( y=x^{3} \)

    c)   \( y= \frac{1}{x} \)

    d)   \( y=5 \)

    e)   \( y=x \)

    Reveal Solution
• , , ,

  to be added

• 0, Chapter 6, Optimisation,

  Optimisation

  Clearly different functions have different values for any given value of the variable. Some functions have a maximum value and others a miniuum value. These are shown in the diagrams below. Other functions have what is know as a point of inflection as also shown below.

  Optimisation is the process of finding the maximum or minimum of a function.

  The maximum and minimum points of any function can be found by working out where the gradient is zero. The points at which the gradient is zero are known as a stationary points. 

  It should also be noted that stationary points can also turn out  to be points of inflection; see below. We have also shown in the diagrams how the sign of the differentiation changes either side of the points (shown in red).

  

  Lets look at some worked examples.

• 1, Chapter 6, Task 1, Worked Example 6,

  Worked Example

  Use the slider to explore worked examples.

  • Worked Example

    A piece of card \(20\;cm\) by \(20\;cm\) has four identical square pieces of side \(x\) removed from the corners so that it forms a net for an open-topped box. The problem is to find the dimensions of a box with the largest (maximum) volume.

    

    Reveal Solution
  • Worked Example

    Find the two stationary points of the function \(T=2k+ \frac{8}{k} \) and determine which is a maximum and which is a minimum.

    Reveal Solution
• , , ,

  to be added

• 0, Chapter 7, Real Life Problems,

  Real Life Problems

  Lets look at how differentiation can be used to solve real life problems.  Here are two worked examples that illustrate such problems.

  Use the slider to explore worked examples.

  • Worked Example 

    You have \(120\) m of fencing and want to make two enclosures as shown in the diagram. The problem is to maximise the area, A, enclosed.

    Use the fact that the total length of fencing is \(\120) m to write an equation connecting \(x\) and \(y\).

    Make \(y\) the subject of this equation and hence write a formula for A in terms of \(x\).

    Now differentiate with respect to \(x\) and put it equal to zero to solve the problem.

    

    Reveal Solution
  • Worked Example 

    A closed cylindrical can has a volume of \(350 cm^{3}\). Find the dimensions of the can that minimise the surface area.

    

    Reveal Solution
• 0, Chapter 8, Summary,

  Summary

  • \(Average\;speed = \frac{total\;distance}{time\;taken}\)
    
  • Instantaneous speed is given by the gradient of a distance-time graph at a particular time
  • The gradient of the quadratic function
    \(y= ax^{2} + bx + c\) is given by \(2ax +b\)
  • Using calculus notation, if \(y = ax^{2} + bx + c\), then its differential is
    \(\frac{dy}{dx} = 2ax + b\)
  • If \(y = x^{3}\), \(\frac{dy}{dx}=3x^{2}\)
  • If \(y=\frac{1}{x}\), \(\frac{dy}{dx} = - \frac{1}{x^{2}}\)
  • The maximum (or minimum) function \(f\left(x\right) \) is given by \(\frac{df}{dx}=0\) (i.e. gradient is zero) 

  

Interactive Exercises:

• Rates of Change , https://www.cimt.org.uk/sif/calculus/c1/index.htm
• Text Exercises, https://www.cimt.org.uk/sif/calculus/c1/text.pdf
• Answers, https://www.cimt.org.uk/sif/calculus/c1/text_ans.pdf

YouTube URL: https://www.youtube.com/watch?v=uD3Sj2wV7SM&list=PL8ce1n6mRlCtS0rO_C7CNwtjhyqDVkr8g&index=21

External Links:

• Rates of Change Video, https://www.bbc.co.uk/bitesize/guides/zpmfgdm/revision/1, images/Images/bitesize-logo.png,
• , , ,

File Attachments: media/course-resources/Rates-Of-Change_Learning-Objectives.pdfmedia/course-resources/Rates-Of-Change_PowerPoint-Presentation.pptxmedia/course-resources/Rates-Of-Change_Text.pdfmedia/course-resources/Rates-Of-Change_Answers.pdfmedia/course-resources/Rates-Of-Change_Essential-Information.pdf